Difference between revisions of "1993 AIME Problems/Problem 2"
(cleanup) |
Mathgeek2006 (talk | contribs) m (→Solution) |
||
| (One intermediate revision by one other user not shown) | |||
| Line 4: | Line 4: | ||
== Solution == | == Solution == | ||
On the first day, the candidate moves <math>[4(0) + 1]^2/2\ \text{east},\, [4(0) + 2]^2/2\ \text{north},\, [4(0) + 3]^2/2\ \text{west},\, [4(0) + 4]^2/2\ \text{south}</math>, and so on. The E/W displacement is thus <math>1^2 - 3^2 + 5^2 \ldots +37^2 - 39^2 = \left|\sum_{i=0}^9 \frac{(4i+1)^2}{2} - \sum_{i=0}^9 \frac{(4i+3)^2}{2}\right|</math>. Applying [[difference of squares]], we see that | On the first day, the candidate moves <math>[4(0) + 1]^2/2\ \text{east},\, [4(0) + 2]^2/2\ \text{north},\, [4(0) + 3]^2/2\ \text{west},\, [4(0) + 4]^2/2\ \text{south}</math>, and so on. The E/W displacement is thus <math>1^2 - 3^2 + 5^2 \ldots +37^2 - 39^2 = \left|\sum_{i=0}^9 \frac{(4i+1)^2}{2} - \sum_{i=0}^9 \frac{(4i+3)^2}{2}\right|</math>. Applying [[difference of squares]], we see that | ||
| − | < | + | <cmath> |
\begin{align*} | \begin{align*} | ||
\left|\sum_{i=0}^9 \frac{(4i+1)^2 - (4i+3)^2}{2}\right| &= \left|\sum_{i=0}^9 \frac{(4i+1+4i+3)(4i+1-(4i+3))}{2}\right|\\ &= \left|\sum_{i=0}^9 -(8i+4) \right|. | \left|\sum_{i=0}^9 \frac{(4i+1)^2 - (4i+3)^2}{2}\right| &= \left|\sum_{i=0}^9 \frac{(4i+1+4i+3)(4i+1-(4i+3))}{2}\right|\\ &= \left|\sum_{i=0}^9 -(8i+4) \right|. | ||
| − | \end{align*}</ | + | \end{align*}</cmath> |
The N/S displacement is | The N/S displacement is | ||
| − | < | + | <cmath>\left|\sum_{i=0}^9 \frac{(4i+2)^2}{2} - \sum_{i=0}^9 \frac{(4i+4)^2}{2}\right| = \left|\sum_{i=0}^9 -(8i+6) \right|.</cmath> |
| − | Since <math>\sum_{i=0}^{9} i = \frac{9(10)}{2} = 45</math>, the two distances evaluate to <math>8(45) + 10\cdot 4 = 400</math> and <math>8(45) + 10\cdot 6 = 420</math>. By the [[Pythagorean Theorem]], the answer is <math>\sqrt{400^2 + 420^2} = 29 \cdot 20 = \boxed{580}</math>. | + | Since <math>\sum_{i=0}^{9} i = \frac{9(10)}{2} = 45</math>, the two distances evaluate to <math>8(45) + 10\cdot 4 = 400</math> and <math>8(45) + 10\cdot 6 = 420</math>. By the [[Pythagorean Theorem]], the answer is <math>\sqrt{400^2 + 420^2} = 29 \cdot 20 = \boxed{580}</math>. |
== See also == | == See also == | ||
| Line 16: | Line 16: | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 10:56, 13 March 2015
Problem
During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went 
miles on the 
day of this tour, how many miles was he from his starting point at the end of the 
day?
Solution
On the first day, the candidate moves ![$[4(0) + 1]^2/2\ \text{east},\, [4(0) + 2]^2/2\ \text{north},\, [4(0) + 3]^2/2\ \text{west},\, [4(0) + 4]^2/2\ \text{south}$](http://latex.artofproblemsolving.com/b/3/2/b324e8eaf2302ed847dd08005526e4104bab8920.png)
, and so on. The E/W displacement is thus 
. Applying difference of squares, we see that
The N/S displacement is
![\[\left|\sum_{i=0}^9 \frac{(4i+2)^2}{2} - \sum_{i=0}^9 \frac{(4i+4)^2}{2}\right| = \left|\sum_{i=0}^9 -(8i+6) \right|.\]](http://latex.artofproblemsolving.com/9/6/6/9669af3dde91ab7da2f1bf716b39f186fcf118bc.png)
Since 
, the two distances evaluate to 
and 
. By the Pythagorean Theorem, the answer is 
.
See also
| 1993 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
